What is the pH of solution obtained by mixing `100 mL` of each `0.2 M NH_(4)Cl` and `0.2M NH_(4)OH`, if `pK_(b)` of `NH_(4)OH` is `4.2`?

Explanation

`pOH = pK_(b) + "log "(["salt"])/(["base"])`
`= 4.2 + log 1 ("because"["salt"]=["base"])`
`= 4.2`
`pH = 14 - pOH = 14-4.2 = 9.8`